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This question already has an answer here:

Suppose the $GCD$ of integers $a$, $b$, and $c$ is 1 when taken in pairs, i.e., they are relatively prime. Prove that $GCD(ab,c)=1$.

Any hints would be appreciated.

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marked as duplicate by Dietrich Burde, user147263, Daniel W. Farlow, Community Feb 24 '16 at 0:50

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    $\begingroup$ Hint: Suppose to the contrary that the gcd is $\ne 1$. Then some prime $p$ divides $ab$ and $c$. Continue. $\endgroup$ – André Nicolas Feb 23 '16 at 19:37
  • $\begingroup$ Alternately, we can use the Bezout "Identity." But that is less natural. $\endgroup$ – André Nicolas Feb 23 '16 at 19:43
  • $\begingroup$ Also, permuting $a,b,c$, see here $\endgroup$ – Dietrich Burde Feb 23 '16 at 20:46
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Suppose gcd$(a,b,c)=d$ where $d > 1$. Then by definition $d|a$ and $d|b$. However, this implies that $d$ is a common divisor of both $a$ and $b$ and since $d > 1$, it contradcits that gcd$(a,b)=1$

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Let A be the set of prime factors of $a$

Let B be the set of prime factors of $b$

Then A$U$B is the set of prime factors of $ab$

If $gcd(ab,c) \neq 1$ then a prime factor of $c$ is also a prime factor of $ab$ and so it belongs to the set $AUB$.

Hence it must also belong to either the set A or B (or both).

Therefore $c$ shares a prime factor with either $a$,$b$ or both, contradicting the fact that $gcd(a,b,c) = 1$

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